End B of rod AB which makes angle θ with floor is being pulled with, a constant velocity v

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3-2.Motion in Plane

hard

The end $B$ of the rod $AB$ which makes angle $\theta$ with the floor is being pulled with, a constant velocity $v_0$ as shown. The length of the rod is $l.$ At the instant when $\theta = 37^o $ then

Avelocity of end $A$ is $\frac{5}{3}$ $v_0$ downwards

Bangular velocity of rod is $\frac{5}{3} \frac{v_0}{l}$

Cangular velocity of rod is constant

Dvelocity of end $A$ is constant

Solution

$V_{A} \cos (90-\theta)=V_{0} \cos \theta$
$\mathrm{V}_{\mathrm{A}} \sin \theta=\mathrm{V}_{0} \cos \theta$
$\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{V}_{0}}{\tan \theta}$
$\mathrm{V}_{\mathrm{A}}=\frac{4}{3} \mathrm{V}_{0}$
$\omega=\frac{\left[V_{A} \sin (90-\theta)+V_{0} \sin \theta\right]}{\ell}$

Standard 11

Physics

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medium

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hard

The end $B$ of the rod $AB$ which makes angle $\theta$ with the floor is being pulled with, a constant velocity $v_0$ as shown. The length of the rod is $l.$ At the instant when $\theta = 37^o $ then

Solution

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